Probability of collecting all the sticker types This question is in the context of tuning a training procedure, whereby the learner may receive random stickers for good performance.  I am trying to figure out the probability of any given learner maximizing their reward by collecting every possible sticker type.  This information will be used to adjust the number of sticker types.
A single learner plays 60 games.  Each game has 100 trials.  Every time the learner succeeds in a trial, there is a 4% chance that the learner will receive a random sticker.  There are an unlimited number of stickers, there are $ n $ sticker types, there is a uniform probability distribution for selecting any given sticker type, and repeat sticker types are possible.  (E.g., Bob won two Batman stickers in a row, but he instead wanted a Batman sticker and a Superman sticker.)
Question: Assuming the learner succeeds in every trial, what is the probability that, at the end of the 60th game, the learner will have collected at least one of each of the $ n $ sticker types?
My thinking thus far:


*

*The problem is equivalent to a single game with 6000 trials.

*The probability of winning any sticker after one trial is $\frac{1}{25}$, or 4%.

*The probability of winning a specific sticker type after one trial is $ \frac{1}{25n} $, or $ \frac{0.04}{n} $.

*$ p_{max} = 1 - p_{fail} $

*The probability of only getting $ i $ unique sticker types after 6000 trials is $p_{i}$.

*$p_{fail} = \sum_{i=0}^{n-1} p_{i}$

*$p_{0} = 0.04^{6000}$ (the probability of not winning any stickers at all after 6000 trials)


At this point I am not sure how to proceed. 
 A: As derived e.g. here, the probability to have all sticker types after obtaining $k$ stickers is
$$
\frac{n!}{n^k}\left\{k\atop n\right\}\;,
$$
where $\left\{k\atop n\right\}$ is a Stirling number of the second kind,
$$
\def\stir#1#2{\left\{#1\atop#2\right\}}\stir kn=\frac1{n!}\sum_{j=0}^n(-1)^{n-j}\binom njj^k\;.
$$
In your case, the number of stickers obtained is binomially distributed, with the probability for $k$ stickers being
$$
\binom{6000}k\left(\frac1{25}\right)^k\left(\frac{24}{25}\right)^{6000-k}\;,
$$
so the probability to have all sticker types is
$$
\left(\frac{24}{25}\right)^{6000}n!\sum_{k=0}^{6000}\binom{6000}k(24n)^{-k}\left\{k\atop n\right\}\;,
$$
which we can transform using the definition of the Stirling numbers:
\begin{align}
&\sum_{k=0}^{6000}\binom{6000}k(24n)^{-k}\frac1{n!}\sum_{j=0}^n(-1)^{n-j}\binom njj^k
\\
={}&\left(\frac{24}{25}\right)^{6000}n!\sum_{j=0}^n(-1)^{n-j}\binom nj\sum_{k=0}^{6000}\binom{6000}k\left(\frac j{24n}\right)^k
\\
={}&
\left(\frac{24}{25}\right)^{6000}n!\sum_{j=0}^n(-1)^{n-j}\binom nj\left(1+\frac j{24n}\right)^{6000}
\\
={}&
n!\sum_{j=0}^n(-1)^{n-j}\binom nj\left(\frac{24}{25}+\frac j{25n}\right)^{6000}\;.
\end{align}
If you don't want to compute this sum, you can approximate the number of stickers obtained by the expected number $6000/25=240$ of stickers obtained, yielding an estimate of
$$
\frac{n!}{n^{240}}\stir{240}n
$$
for the probability to have all sticker types.
